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I am trying to understand fully how drug half-life works. So I derived this relationship:

$$\ U_{r} = \frac{1+\ U_{r-1}}{2}$$ Where $\ U_{0}=0$ and r is a set of natural numbers.

My issue to how to deduce a relationship for the sum to infinity:

$$\ S_{\infty}=\lim_{n\to\infty} \sum_{r=1}^n \ U_{r}$$

Consequently I need to get the relationship for $\ S_{\infty}$ if $\ U_{r} = \frac{A+\ U_{r-1}}{2}$ and $\ U_{0}=0$

Obinoscopy
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    The recurrence is easily solved: $U(r)=A 2^{-r} \left(2^r-1\right)$. More problematic is that this approaches $A$ as $r\to\infty$, which means unless $A=0$ in which case the expression is trivial, the sum doesn't converge. – Integrand Jun 11 '20 at 15:05
  • You can solve the recurrence like this \begin{align} U_r & = \frac{1}{2} + \frac{U_{r - 1}}{2} = \frac{1}{2} + \frac{\frac{1 + U_{r - 2}}{2}}{2} = \frac{1}{2} + \frac{1 + U_{r - 2}}{4} = \frac{3}{4} + \frac{U_{r - 2}}{4} = \frac{3}{4} + \frac{\frac{1 + U_{r - 3}}{2}}{4} \ & = \frac{7}{8} + \frac{U_{r - 3}}{8} = \ldots\end{align} What happends if you do this down to $r = 0$? – ViktorStein Jun 11 '20 at 15:09

2 Answers2

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$U_r=\frac{1}{2}+\frac{U_{r-1}}{2}=\frac{1}{2}+\frac{1}{4}+\frac{U_{r-2}}{4}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{U_{r-3}}{8}=...=\sum_{i=1}^{r}\frac{1}{2^i}+\frac{U_0}{2^r}=1-\frac{1}{2^r}$

and hence $\sum_{1}^{n}U_r=n-\sum_1^n\frac{1}{2^r}$

aritracb
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Suppose $u_0,u_1,u_2,...$ are defined by $$ \left\lbrace \begin{align*} u_0\!&=\,0\\[4pt] u_{n+1}\!&=\frac{a+u_n}{2}\\[4pt] \end{align*} \right. $$ for some constant $a > 0$.

Then we have \begin{align*} u_1&=\frac{a}{2}+\frac{1}{2}u_0\\[4pt] u_2&=\frac{a}{2}+\frac{1}{2}u_1\\[4pt] u_3&=\frac{a}{2}+\frac{1}{2}u_2\\[4pt] \!\!\!\!\!\vdots&\phantom{=\frac{a}{2}}\;\;\,\vdots\\[4pt] u_{n}&=\frac{a}{2}+\frac{1}{2}u_{n-1}\\[4pt] \end{align*} Summing the above equations, we get $$\;\;\;\;S_n=\frac{a}{2}{\,\cdot\,}\,n+\frac{1}{2}S_{n-1}$$ but it's easily shown that for all $k$ we have $S_k\ge 0$, so $$S_n\ge \frac{a}{2}{\,\cdot\,}\,n\qquad\;\;$$ hence ${\displaystyle{\lim_{n\to\infty}}}S_n=\infty$.

quasi
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